Question

The first and the last terms of an $A.P.$ are $a$ and $l$ respectively. Show that the sum of nth term from the beginning and nth term from the end is $a+l$.

Answer 1

$a_1=a$

Let

$ak=l$

$a+(k-1)d=l$

$d=\frac{l-a}{k-1}$

$n^{th}$ term from beginning $=a_n$

$=a+(n-1)d=a+(n-1)\left(\frac{l-a}{k-1}\right)$

$n^{th}$ term from end $=l+(n-1)(-d)=l-(n-1)\left(\frac{l-a}{k-1}\right)$

Sum of $n^{th}$ terms from beginning and from end $=a+(n-1)\left(\frac{l-a}{k-1}\right)+l-(n-1)\left(\frac{l-a}{k-1}\right)=a+l$

proved.